What is the integral of 1/(36+9x^2) ?

The integral of 1/(36+9x^2) is 1/18 arctan(x/2)+C

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integral of 1/(36+9x^2)

\int \frac{1}{36 + 9 x ^{2}} d x

\frac{1}{18} arctan (\frac{x}{2}) + C

Solution steps

\int \frac{1}{36 + 9 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{18 ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{18} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= \frac{1}{18} arctan (u)

Substitute back

u = \frac{3}{6} x

= \frac{1}{18} arctan (\frac{3}{6} x)

Simplify

\frac{1}{18} arctan (\frac{3}{6} x) : \frac{1}{18} arctan (\frac{x}{2})

= \frac{1}{18} arctan (\frac{x}{2})

Add a constant to the solution

= \frac{1}{18} arctan (\frac{x}{2}) + C

\int sec^{2} (x) sin (tan (x)) d x

\int \frac{3 x ^{2}}{sec ( 2 x )} d x

\int e^{2 x - y} d x

\int x^{2} + 15 d x

9 - x^{2}

What is the integral of 1/(36+9x^2) ?
The integral of 1/(36+9x^2) is 1/18 arctan(x/2)+C